Determining the result or state of system in a calorimeter

[Moose100]

1. The problTfem statement, all variables and given/known data
Determine the result when 100g of steam at 100C is passed into a mixture of 200g of Water and 20g of ice at exactly 0C in a calorimeter which behaves thermally as if it were equvalent to 30g of water.

Homework Equations

FInding the mass. I think I figured out the final temp?

The Attempt at a Solution

Q=0=Steam condensed heat + Change in steam water heat + change in water heat + heat to melt Ice + change in calorimeter heat

(540cal/g)(100g) + (100g)(1cal/g C)(Tf-0C) +(200g)(1cal/g C)(Tf-0C) +(20g)(80g/C) +(20g)(1cal/g C)(Tf-0C) + (30g)(1cal/gC)(Tf-0C)

Tf=149.7

ITs all water so the temp has to be 100 it can't go higher than that.I am struggling to find the grams of steam condensed.

m(-540g/C)=Ice melted + Rise in water?

Please help and thanks!

 

[collinsmark]

1. The problTfem statement, all variables and given/known data
Determine the result when 100g of steam at 100C is passed into a mixture of 200g of Water and 20g of ice at exactly 0C in a calorimeter which behaves thermally as if it were equvalent to 30g of water.

Homework Equations

FInding the mass. I think I figured out the final temp?

The Attempt at a Solution

Q=0=Steam condensed heat + Change in steam water heat + change in water heat + heat to melt Ice + change in calorimeter heat

(540cal/g)(100g) + (100g)(1cal/g C)(Tf-0C) +(200g)(1cal/g C)(Tf-0C) +(20g)(80g/C) +(20g)(1cal/g C)(Tf-0C) + (30g)(1cal/gC)(Tf-0C)

Tf=149.7

ITs all water so the temp has to be 100 it can't go higher than that.

I got a different numerical answer. But yes, I agree (even with my answer) that it was above 100, so we must use a different approach where the final state is steam and water.

I am struggling to find the grams of steam condensed.

m(-540g/C)=Ice melted + Rise in water?

Yes, that's the right approach.

Set up an equation and see what you get.

(As you've been doing already, continue to remember that once the ice melts it creates additional water. Also in this case, the calorimeter acts like water. You've accounted for these things before; just be sure to continue that trend when you set up your new equation.)

Edit: btw, the manner in which you are expressing units is a little off. Heat of vaporization has units of cal/g, not g/C.

 

[insightful]

It's a lot easier to solve in stages:

1. Melt ice.
2. Heat up to (max) 100C.
3. What happens to any excess steam (assumes atmospheric pressure calorimeter).

 

[Moose100]

I got a different numerical answer. But yes, I agree (even with my answer) that it was above 100, so we must use a different approach where the final state is steam and water.
Yes, that's the right approach.

Set up an equation and see what you get.

(As you've been doing already, continue to remember that once the ice melts it creates additional water. Also in this case, the calorimeter acts like water. You've accounted for these things before; just be sure to continue that trend when you set up your new equation.)

Edit: btw, the manner in which you are expressing units is a little off. Heat of vaporization has units of cal/g, not g/C.

Oops typed tht wrong

 

[Moose100]

I got a different numerical answer. But yes, I agree (even with my answer) that it was above 100, so we must use a different approach where the final state is steam and water.
Yes, that's the right approach.

Set up an equation and see what you get.

(As you've been doing already, continue to remember that once the ice melts it creates additional water. Also in this case, the calorimeter acts like water. You've accounted for these things before; just be sure to continue that trend when you set up your new equation.)

Edit: btw, the manner in which you are expressing units is a little off. Heat of vaporization has units of cal/g, not g/C.

SO basically it's the same setup as above but with m "missing"

 

[collinsmark]

SO basically it's the same setup as above but with m "missing"

Basically yes. Although it's really even a tad simpler.

This time around, you know what the final temperature is.

Also, this time around, you don't need to worry about the portion of water, which was once steam, dropping to some final temperature. The Final temperature is 100 ^oC. Yes, there is steam that condenses into water, but it doesn't change temperature from there. So that simplifies things.

 

[Moose100]

Basically yes. Although it's really even a tad simpler.

This time around, you know what the final temperature is.

Also, this time around, you don't need to worry about the portion of water, which was once steam, dropping to some final temperature. The Final temperature is 100 ^oC. Yes, there is steam that condenses into water, but it doesn't change temperature from there. So that simplifies things.

In other words that term goes to zero?

 

[collinsmark]

In other words that term goes to zero?

Yes. For that particular term, the water has a temperature change of 0. So the term goes to 0, but just that particular term.

 

[Moose100]

Right that's what I meant.

 

[Moose100]

Yes. For that particular term, the water has a temperature change of 0. So the term goes to 0, but just that particular term.

SO a few minor questions since we solved it. THe first equation describes the system

Like why does it work afterword to find grams of steam? Is it because one doesn't exhibit a change? The change in temp of steam water? These are basically before and after states or equations?

 

[Moose100]

IM running into problems with the same type of problem with different numbers

Mass steam 10g at 100C
Mass Water at 0C
Mass Ice at 0C
Mass Calorimeter at 0C(equivalent to water.

Same thing except I get -2.5C. for a final temp.

It also looks at ice meled instead of steam condensed. How do you know which one to pick?

 

[Moose100]

Hmm.

When we look at hese final systems in both instances how do we know what to pick? IN terms of ice or steam mass to find?

In the first case the final temperature is 149C or so. SInce it is water it can't go higher. so it is really 100C. How do we know to analyze steam? Does it mean that the stuff in the calorimeter GAINED some energy? From the steam since there was so much?

In the second the negative number means that the system or the water LOST energy to the ice? Melting it?

Its like depending on how much you have one steam or ice buffers the temp? THanks

I feel like conceptually I am still quite not getting it I have a feeling it depends on what that final temperature number means in all three states of water in the system and who is exchanging with what...

 

[Moose100]

Ahh I found it.

WHat I need to do is look in each case are the energies needed to metl ice and condense steam. For instance in the the first example the energies for the ice and steam are 1600 cal and 54000 cal respectively. Also to raise the contents(calorimeter mass + water + ice water mass) with steam takes 250g(100C)(1cal/gC) =26,000cal(this is the energy at equilibrium the ice would have to melt so it would all be water) Since the steam is way larger than the system all of it melts and manages only to condense some of the steam raising the while raising the water . This means that we can find out how much steam is condensed:

m(-540g/C)= 26,600. m=49.

The way I ws doing it before made it difficult to "pick" which one I needed to find the mass for. We can get the final temp but I don't know what it means for each unless I could quantitiatively compare the energies of each component(Contents of total calorimeter ice vs steam.).

"Ice vs steam" tells me which one will be left over. Total calorimeter tells me what final engery we have to find the mass of said state of water left over.

Unless someone can tell me how to do so with the way I did it BEFORE this?

THanks.

 

[Moose100]

Can someone tell me if i have the right approach I am unsure still how to approach these in general.

 

[collinsmark]

I think you might have answered your own questions.

But elaborating, it all comes down to conservation of energy.

• The energy of the steam condensing must be equal to the energy of the ice melting, if the final state is ice and water.
• The energy of the steam condensing must be equal to the energy of the ice melting, plus the energy of the water (some of it previously ice) and calorimeter changing temperature, if the final state is all water (and calorimeter), where the final temperature is in between 0 and 100 deg C.
• Same as above if the final temperature is 100 deg C, except in this case, not all the steam condenses.

Conservation of Energy applies in any case.

So how do you which one is the final state? Well, there's a couple of ways. You could try it one way and see if the results make sense. If they don't try it a different way. Or, you could calculate the energies of certain, particular changes and compare them, for example comparing the energy of the total steam condensing compared to the energy of the ice melting. That can at least help narrow down the possibilities.

 

[Moose100]

My approach I think is I will compare the energies of steam and ice so like in the case of the first the steam is larger. That allows me to identify that the system is heating up to 100C(stuff in calorimeter) and that since one is larger the ice melts. This means that essentially the final temp is 100.
SInce Qlost=Qgained:

M(540g/C)= basically the rest of the systems compontents.

The one that exhibits no change in temp is the steam water itself and that term will go to zero.

with a change in temp of 100 degrees plus the energy of ice melting it comes to equaling 26600 cals

so mass is what we said before

This works for the other case when the system COOLS the ice wins eats up all the steam and we can the amount of ice melted.

THis makes more sense to me because it depends on which way we are traveling on the graph or trend. It also allows me to remember that steam(ice) and water can exist at 100C(0C for ice).So identify what happens to ice and steam. Then assess if heating or cooling to a final temp then solve accordingly.

I came to this awhile ago and was ecstatic!

 

[collinsmark]

Wonderful!

For what it's worth, the answer I came up with is also 49 g of steam condensed (leaving [100 - 49] g of steam not condensed). (Don't forget your units. )

 

[Moose100]

Bah! Who needs units you know what I mean! :p

This method also works for post 11. This allows me to identify and track what's going on.

 

[Moose100]

Track after going in blind